题目：
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The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row. The goal is to take cards in such order as to minimize the total number of scored points. For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring 10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000 If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be 1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150. |
input
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The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces. |
output
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Output must contain a single integer - the minimal score. |
sample
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6 10 1 50 50 20 5 |
Sample Output
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3650 |
题目大意：
一个整数序列包含N个1~100的整数（3<=N<=100），从中取出一个数并和相邻两边的整数相乘，依次进行下去直到只剩下首尾两个数为止，求最终的得到的和的最小值。两边的数不能取，且不重复选取。
思路：
dp[i][j]表示将第I个和第j个之间的数取完得到的和，那么由k表示的数决定的子序列的宽度，依次递增宽度，并且移动子序列，即改变i的值，然后对i到i+k之间的元素进行遍历，比较dp[i][i+k]和先选取a[i]和a[j]之间的数，a[j]和a[i+k]之间的数，最后只剩下a[j]，这一过程中的得到的和dp[i][j]+dp[j][i+k]+a[i]*a[i+k]*a[j]与原来的和的大小。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include<cstdlib>
#include<ctype.h>
using namespace std;
typedef long long ll;
#define pi 3.14159265358979323846264338327
#define E 2.71828182846
#define INF 0x3f3f3f3f
#define maxn 555555
int a[110];
int dp[110][100];
int main() {
int T;
while (scanf("%d", &T) != EOF) {
int i, j, l;
for (i = 1; i <= T; i++) {
scanf("%d", &a[i]);
}
memset(dp, 0, sizeof(dp));
for (l = 2; l < T; l++) {
for (i = 2; i + l <= T + 1; i++) {
j = i + l - 1;
dp[i][j] = INF;
for (int k = i; k < j; k++) {
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + a[i - 1] * a[k] * a[j]);
}
}
}
printf("%d\n", dp[2][T]);
}
return 0;
}
原文地址：https://www.cnblogs.com/ineedyou/p/8456058.html